1. consider both ± ∞
as x is a huge negative number the x^2 would make it positive, but very large, multiplying by -5 makes it "hugely" negative. As x is large, both ±, the -3x+1 becomes insignificant.
so for large + x's the same thing happens, the function becomes "hugely" negative.
so the parabola drops down into the third quadrant and down into the fourth quadrant.
Is that what you want ?
for #2
you have 2/(-4x^2)= -1/(2x)^2
should have been -1/(2x^2)
you have the exponent of 2 outside the brackets, thus also squaring the 2 of the 2x, only the x is squared.
1)Without graphing, describe the end behavior of the graph of f(x)= -5x^2-3x+1
as x starts from a negative value and increases the equation -5x^2-3x+1 curves up and peaks and comes down and passes the y-intercept at (0,1) to negative infinity.
I don't know what the answer is. I got it wrong, I would love some el helpo
2)If f(x)= -4x^2 and g(x)= 2/x. Find [g0f](x)
g(F(x))=g(-4x^2)= 2/(-4x^2)= -1/(2x)^2
-1/(2x)^2 = [g0f](x)
This is wrong too, maybe it's suppose to be 1/(-2x^2)? I don't really see what the difference is.
2 answers
I guess that's what they want for #2 Thank you so much
1 answer